Equation 2.6.1 is correct.
Don't forget that in mesh analysis we can choose current direction arbitrary.
The author choose clockwise direction and by doing this he clearly defined polarities (the voltage drops) across resistors in the circuit according to the assumed directions of the mesh currents.

And this is why KVL looks like this

-12V + 4Ω*I + 2*Vo - 4V + 6Ω*I1 = 0

And Vo do not show the voltage polarities across 6Ω resistor.
This Vo show as how our VCVS will "sample/see" the voltage across 6Ω resistor.
And for our assumed current direction VCVS will "sample" negative voltage.
And this is why Vo = -6Ω*I

And for example if we assume counterclockwise direction we have this situation

And KVL

4V - 2*Vo + 4Ω*I + 12V + 6Ω I = 0

And

Vo = 6Ω*I

And finally we can use third method.

-12V + 4Ω*I + 2*Vo - 4V - Vo = 0

And Vo = -6Ω*I is negative because the Vo polarity is opposite to the direction of the loop current.